What Is the Minimum Force Needed to Push a Tall Dresser Across the Floor?

[amcavoy]

Say you have a dresser that is to be pushed across a floor. The height is x, the width is y, and the coefficient of static friction is u. If you apply a force F and the top left corner (height x), what is the minimum value of F so the the dresser will move across the floor?

I assumed here that the center of gravity is in the middle of the dresser somewhere. My approach to this problem would be then to calculate the torque from the force F and the torque from the torque from the friction. However, I don't really know how to factor in the width of the dresser. Does anyone have any ideas on this?

Thank you.

 

[LeonhardEuler]

Well, suppose the force you exert is perfectly horizontal. Then there are four forces acting on the dresser that we are concerned with: Your pushing force, the friction force, gravity and the normal force. Balancing forces in the vertical direction tells you that the normal force is equal in magnitude to the force of gravity, mg. Now, to consider when the dresser either starts sliding or tipping over, assume the force of friction is at a maximum, $F_f=\mu N=\mu mg$. Balancing forces in the horizontal direction will tell you that the force you must exert is $F_{push}=\mu mg$. But this does not take into account the torques. When you start pushing an object the pushing force and friction force form an action reaction pair that generates a torque. As the object shifts slightly, one side of the object is pressed to the ground more forcefully than the other. This recenters the normal force. This causes the gravitational and normal forces to create a counter-torque. The object will tip over once the normal force needs to be centered ouside of the object balance the pushing and friction torque.